What is the demand load in amps for a range with a rating of 6 kw single phase?

Answer 1

Watts (Power, P) are the product of volts (V) and amps (I). For single-phase loads: P = IV, P/V = I, P is 17.7kW = 17700W, V is 240, 17700W/240V = 73.75A The rule of thumb for breaker size is you want your expected load to be no more than 80% of your breaker size. Or, put another way (the way the code states), your circuit must be sized to handle 125% of the connected load. Two ways of saying the same thing. 73.75A/80% = 92.1975A You will need a 100A breaker for this range. You should consult an electrician for connecting this; it is a very heavy load. This is probably not worth trying to install yourself. You will need at least #2 copper or #1 aluminum wire, and wire that heavy is not fun to work. ADDITIONAL: The above answer is not correct for the US. It IS correct for general loads, but not for ranges.It may apply in other countries (I dont know for sure whether it does or not). In the US at least, the NEC allows for a very different figure to be used for electric cooking appliances. The question uses the specific term "demand load", suggesting this is actually a code question. Consult NEC-2005 Table 220.55 notes 1 and 4. For a single cooking appliance rated over 12kW and under 27.5kW you take the demand factor for a 12kW range and add 5% for each kw or fraction thereof over 12. So, for this example: The demand factor for a 12kW range is 8kW Our example range draws 5.7 kW more than 12 so we have to add 30% (5% X 6kW). The additional kW is only 5.7, but the code says we have to treat the 0.7 fraction as the next whole number, thus 6kW instead of 5.7. 0.3 (30%) X 8 (base kW) = 2.4kW 8kW + 2.4kW = 10.4kW 10.4kW / 240V = 43.3A The demand load in amps for this range (in the US) would be 43.3A. The premise of the code is that the oven and all of the stove burners would never be on at the same time. It would be perfectly legal and safe in the US to install this range with a 50 amp breaker. You can use #8 copper wire as long as the wire and connection points are rated for 75 degreesC, or #6 copper if the ratings are 60 degreesC. There is one exception. If you use Romex (NM-B), the code says you must use the 60 degreesC table, so you would have to use the #6, even though NM-B is rated at 90 degreesC. Go figure. NEC article 422.10(A) exempts the branch circuit feeding the range from the 80% rule. It says the branch circuit rating for household cooking appliances shall be determined in accordance with 220.55. As you can see, If you don't have the proper training, you would have purchased a breaker and wire that cost way too much, and struggled with Wrestling it in. The inspector may or may not have approved it. Consult a pro for at least the design calcs! It will save you time and money in the long run.

Answer 2

It is found by dividing the power in watts by the voltage, so 6000 divided by your supply voltage. For a 240 v supply, that is 25 amps.

Answer 3

About 33 amps use a 40 amp breaker.